P is properly contained in EXP

A fixed transducer reads $1^n$, counts $n$, finds $k=\lceil \sqrt{n}\rceil$ by integer search for the least $k$ with $k^2\ge n$, and writes the binary word $1$ followed by $k$ zeros. For the finitely many lengths where $n+1>2^k$, a literal branch instead writes $\mathrm{bin}(n+1)$. Its time is polynomial in $n$ plus $O(k)$, hence $O(f(n))$, since every fixed polynomial in $n$ is eventually below $2^{\sqrt{n}}$. A second transducer reads $1^n$ and writes $1$ followed by $n$ zeros in $O(n+1)=O(2^n)$ time. Both fully read the input and $f,g\ge n+1$, so they are time constructible in the stated output sense. Every polynomial is $O(f)$ because $d{\log }_{2}n=o(\sqrt{n})$ for each fixed exponent $d$, and $f\log f=o(g)$ because $\sqrt{n}+{\log }_2(\sqrt{n}+1)=o(n)$.

The deterministic time hierarchy applied to step 1.1 gives $\mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(f)⊊\mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(g)$. Since $P\subseteq \mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(f)$ and $\mathrm{D}\mathrm{T}\mathrm{I}\mathrm{M}\mathrm{E}(g)\subseteq EXP$, it supplies a language in $EXP\setminus P$.